My image is broken when im displaying using php - php

Question

When im displaying the image of restaurant but the image is broken.can anyone help me to solve or find out the problem?i tried to googling or looking tutorial yet i found there are several way to store image which is either storing the file path to database or storing the image to database,in my case is storing the file path to database.
this is the data in my database.the image is type is blob.
this is my html code for displaying
<?php
$sql = "SELECT * FROM restaurant";
$result = $conn->query($sql) or die($conn->error);
while ($row = $result->fetch_assoc()){
<tr>
<td>{$row['ID']}</td>
<td>";echo "<img src='".$row['Logo']."' />";echo "</td>
<td>{$row['Name']}</td>
<td>{$row['ContactNumber']}</td>
<td>{$row['Address']}</td>
<td>{$row['CuisineType']}</td>
<td>{$row['SupportArea']}</td>
</tr>\n";
}
?>
this is php code for adding restaurant
<?php
include 'db.php';
//define other variables with submitted values from $_POST
$name = $_POST['restaurant_name'];
$contactnum = $_POST['restaurant_contactnum'];
$address = $_POST['restaurant_address'];
$cuisinetype = $_POST['cuisinetype'];
$checkbox1 = $_POST['area'];
$allowedExts = array("png","jpg","jpeg"); /* ACCEPTED FILE FORMAT */
$extension = pathinfo($filename, PATHINFO_EXTENSION); /* GET THE FILE EXTENSION */
$extension = strtolower($extension); /* LOWER THE STRINGS OF THE EXTENSION */
$chk="";
foreach($checkbox1 as $chk1)
{
$chk.= $chk1.",";
}
if(in_array($extension,$allowedExts)){ /* IF FILE IS INDEED AN IMAGE */
$photo="restaurantlogo/" . $_FILES["restaurant_logo"]["name"];
}
//insert user data into database
$sql = "INSERT INTO restaurant (Logo, Name, Password, ContactNumber, Address, CuisineType, SupportArea ) VALUES ('$photo', '$name', '$password', '$contactnum', '$address', '$cuisinetype','$chk')";
$insert = mysqli_query($conn, $sql) or die(mysqli_error($conn));
//check if mysql query is successful
if ($sql){
//redirect the user to welcome.php
header("Location: ../restaurant-add.php");
}

Answer 1

The problem is in your PHP output. You were missing echo at multiple places and I exchanged the " for ' so the HTML <img>'s src tag is wrapped in " ".
Here is the corrected code:
<?php
$sql = "SELECT * FROM restaurant";
$result = $conn->query($sql) or die($conn->error);
while ($row = $result->fetch_assoc())
{
echo '<tr>
<td>'.$row['ID'].'</td>
<td><img src="'.$row['Logo'].'" /></td>
<td>'.$row['Name'].'</td>
<td>'.$row['ContactNumber'].'</td>
<td>'.$row['Address'].'</td>
<td>'.$row['CuisineType'].'</td>
<td>'.$row['SupportArea'].'</td>
</tr>";
}
?>

Answer 2

Click F12 to open inspect element, check if you are getting valid src path for image.
If it is valid, try to open in in new tab and see if it appears there. Clearly you will find your error there